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Why $\lim_{x \to ∞} 1^x=1$ even though $1^∞$ in indeterminate? I'm genuinely not getting it.

Any help is appreciated!

EL02
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  • Because $1^x$ is constant equal to $1$. This is the same as $0 \times x$, which is constant equal to $0$ and is not an indeterminate form $0 \times \infty$. – TheSilverDoe May 04 '21 at 09:00
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    “Indeterminate form” doesn't mean “we can't know what the result is”. – egreg May 04 '21 at 09:01
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    It's best to not really think about expressions like $1^{\infty}$ or $0 \cdot \infty$, etc. The symbol $\infty$ is not a number, computing a limit is computing which value an expression approaches. – Qi Zhu May 04 '21 at 09:04
  • @TheSilverDoe but then again why $\lim_{n \to ∞} (\frac {n}{n+1}) ^n ≠1$ even though it is equal to $1^∞$? – EL02 May 04 '21 at 09:14
  • @GreasyLlama Apply the definition of the limit and you will see. Also read the answer of J. Darne to understand why the form $1^\infty$ is meaningless ($\sim$ indeterminate). – user May 04 '21 at 09:17

2 Answers2

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Well, the fact that $1^\infty$ is indeterminate does not mean that $f(x)^{g(x)}$ has no limit when $f(x)$ approaches $1$ and $g(x)$ approaches $\infty$. It only mean that you cannot deduce the limit of $f(x)^{g(x)}$ from the information "$f(x)$ approaches $1$ and $g(x)$ approaches $\infty$". For some $f$ and $g$ satisfying this hypothesis, the limit can exist, and be $1$, or $4$, or $\pm \infty$, or anything, or not exist. It turns out that for $f$ constant equal to $1$, $f^{g}$ is constant equal to $1$, and the limit is $1$. But the fact that you can compute the limit in this case does not contradict the indeterminacy statement.

J. Darné
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Note that $$ 1^x = \mathrm{e}^{x\ln 1} = \mathrm{e}^0 = 1 $$

for each $x\in \mathbb R$. Hence, the function is constantly $1$ even before doing the limit.

Sewer Keeper
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